Question 756385
solve the following equation for all angles in [0,2pi) 
sin(2x-pi/3)= square root 3/2
2x-π/3=π/3, 2π/3 (in Q1 and Q2 where sin>0)
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2x-π/3=π/3
2x=2π/3
x=π/3
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2x-π/3=2π/3
2x=3π/3=π
x=π/2